Chapter 10 Electricity and Magnetism

A portion of a circuit is shown below. Is the current through the 12 ohm resistor is 1A, what is the value of current I? Resistor is in parallel. R1 = 12 ohm, R2 = 4 ohm, current through R1 = 1 A.

(12 x 4) / (12 + 4) = 3 ohm Top voltage drop: V = IR V = 1A x 12 ohm = 12V Bottom: I = V/R I = 12 V / 4 ohm = 3A 1A + 3A = 4A

What current can cause death?

.2A

Dielectric constant of air

1

What are the two purposes of a parallel plate capacitor?

1. Create a uniform electric field (points to - plate) 2. Store electric potential energy

Capacitors in series equation

1/C=1/C₁+1/C₂+...

Equivalent resistance in parallel

1/Rp=1/R1+1/R2+1/R3+...+1/Rn

A portion of a circuit is shown below. If the current entering the parallel combination is 12A, how much current flows through the 120 ohm resistor. R1 and R2 arre in parallel, R1 = 120 ohm, R2 = 60 ohm

x + 2x = 12 3x = 12 x = 4A I top = 4A I bottom = 8A V = IR V = (4A)(120 ohm) V = 480 V V = IR V = (8A)(60 ohm) V = 480 V (this should match the first value)

The ground is considered to be a potential of _______

zero

What is the ideal resistance of an ammeter?

zero

What is a parallel plate capacitor?

A parallel plate capacitor consists of two parallel plates separated by an insulator. This insulator is called a dielectric.

The diagram below shows a battery with an emf of 100V connected to a circuit equipped with a switch, S. A. What is the current in the circuit when the switch is open When open, it forms a resistor in series (R1 + R2) R1 = 80 ohm R2 = 120 ohm B. What is the current in the circuit when the switch is closed? When open, R1 and R2 are in series and in parallel with R3 R3 = 50 ohm

A. V = IR I = V/R I = 100V/200 ohm = .5A B. 200 x 50 / (250) = 40 ohm V = IR I = V/R I = 100/40 I = 2.5 A

A capacitor with no initial charge is connected in a circuit to a battery and allowed to charge fully. Which of the following best describes the charging process of the capacitor? A. The work done by the battery becomes the electric potential energy stored on the capacitor. B. The work done by the capacitor becomes the electric potential energy stored on the capacitor. C. The work done by the battery becomes the electric kinetic energy stored on the capacitor. D. The work done by the capacitor becomes the electric kinetic energy stored on the capacitor.

A; Choice B is false because a system (in this case the capacitor) doing work reduces its potential energy. Choices C and D are false because kinetic energy cannot be stored (stored energy is potential energy, whereas kinetic energy is energy of motion). Once the capacitor begins charging, extra electrons are being removed from the positively charged plate and being moved to the negatively charged plate. This work against the electric field is stored by the capacitor as electric potential energy. The outside agent doing this work is the battery.

A student in physics lab creates a simple particle accelerator using a parallel plate capacitor. The top plate has a positive charge and the bottom plate has a negative charge, the plates are separated by a distance d, and a uniform electric field is created between the plates. The student can adjust the voltage differential V on the plates and is testing particles of different masses m and charges q. Which of the following shows the calculated acceleration of a particle due to the electric field, based on known quantities? A. qV / md B. md / qV C. Vd / mq D. qVd / m

A; The acceleration comes from the electric force on the particle due to the electric field. The electric force is F = qE where E is the strength of the electric field. The electric field and potential difference on a capacitor are related by the equation V = Ed so E = V / d and F = qV / d. Using Newton's second law F = ma yields ma = qV / d and a = qV / md. F = qE V = Ed F = qv/d ma = qv/d a = qv/md

A test charge 2 nC and mass 10-8 kg is released from rest at point A which has an electric potential of -100 V and travels to point B which has an electric potential of -500 V. What is the speed of the test charge when it reaches point B? A. 12.6 m/s B. 16 m/s C. 80 m/s D. It cannot be determined from the information given.

A; The change in potential energy can be calculated using ΔPE = qV = (2 × 10-9)(-500 - -100) = -8 × 10-7. Since this represents a loss in potential energy, there must be a gain in kinetic energy of the same amount. Both the initial speed and kinetic energy are zero since the charge started at rest, so the entire change in kinetic energy is due to the new speed. Therefore KE = (1/2) mv2 and v2 = 2KE / m = 2(8 × 10-7) / 10-8 = 160 and v = √160 ≈ 13 m/s. Note that the exact square root is not needed since only choice A is close.

With the information given below, what is the voltage of the battery? R1 = 20 ohm, R2 and R3 are in parallel, R2 = 5 ohm, R3 = 10 ohm, and the current through R3 = 3A A. 150 V B. 210 V C. 240 V D. 300 V

B 1. V = IR V = (3)(10) V = 30V 2. V = IR I = V/R I = 30 V/5ohm I = 6A 3. 3A + 6A = 9A 4. V = IR V = (9A)(20 ohm) V = 180 V 5. 180 V + 30 V = 210 V

Consider two copper wires. Wire 1 has three times the length and twice the diameter of wire 2. If R1 is the resistance of wire 1 and R2 is the resistance of wire 2, which of the following is true? A. R2 = (2/3)R1 B. R2 = (4/3)R1 C. R2 = 6R D. R2 = 12R

B L1 = 3L2 d1 = 2d A1 = 4A2 R2/R1 = pL2A2/pL1A1 R2/R1 = L2A2/L1A2 R2/R1 = (1/3)(1/4) R2/R1= 1/3 * 4/1 R2 = 4/3 R1

After a hot, dry day, clouds build in the sky as a thunderstorm approaches. The charge build up on a cloud is 40 C and takes roughly 2 hours to build. If the instantaneous current of a lightning bolt is 30 kA, how long does it take for a lightning bolt to discharge all the charge built up in the cloud? A. 0.75 ms B. 1.3 ms C. 8.3 ms D. 11.1 ms

B; Current is defined as charge per time, or I = Q / t. Solving for time yields t= Q / I = 40 / (30 × 103) = 1.3 × 10-3 s = 1.3 ms.

A fully charged capacitor is connected in a circuit to a light bulb and allowed to discharge fully. Which of the following best describes the discharging process of the capacitor? A. The work done by the capacitor becomes the electric potential energy stored on the light bulb. B. The work done by the capacitor becomes the kinetic energy of the charges, and the light and heat energy of the light bulb. C. The work done by the light bulb becomes the electric potential energy stored on the capacitor. D. The work done by the light bulb becomes the kinetic energy of the charges, and the light and heat energy of the capacitor.

B; Once the capacitor begins discharging, extra electrons are being removed from the negatively charged plate and being moved to the positively charged plate. This work from the capacitor is causing the charges to move (kinetic energy) through the circuit and light bulb. In the light bulb, the motion of the charges creates light and heat.

A parallel plate capacitor with a capacitance of C is connected to a battery with a voltage V and allowed to charge fully to a charge of Q. If the area of the plates on the capacitor is doubled and the capacitor is again allowed to charge fully, all of the following are true EXCEPT: A. C doubles B. V doubles C. Q doubles D. The potential energy stored on the capacitor doubles

B; The capacitance of a capacitor is directly proportional to the area of the plates, so doubling the area will double the capacitance, eliminating choice A. When the capacitor is allowed to charge fully, the plates will have the same voltage differential as the battery, so V will remain the same regardless of the area of the plates, making choice B false and the correct answer. Since V is not changing and C is doubling, then by Q = CV the charge Q must double, eliminating choice C. Since potential energy is (1/2) QV and Q doubles while V stays the same, then the potential energy must also double, eliminating D.

A capacitor is connected in a circuit to a battery and allowed to fully charge. The battery is disconnected. Then a dielectric is inserted between the plates of the capacitor. If the dielectric constant is K, all of the following are true EXCEPT: A. The charge stored on the capacitor remains the same. B. The potential energy stored on the capacitor increases by a factor of K. C. The voltage across the capacitor decreases by a factor of K. D. The electric field created by the capacitor decreases by a factor of K.

B; When the dielectric is inserted, the capacitance of the capacitor increases by a factor of K. Since the battery is not connected, the charge stored on the plates will not change, eliminating choice A. Using Q = CV with Q the same and C increasing by a factor of K, then V must also decrease by a factor of K, eliminating choice C. Using PE = (1/2) QV then PE must also decrease by a factor of K, making choice B false and the correct answer. Using V = Ed with V decreasing by a factor of K, then E must also decrease by a factor of K, eliminating choice D.

What will happen to the capacitance of a parallel plate capacitor if the plates were moved closer together, halving the distance between them?

C = E0/ (.5d) C would double

How big would the plates of a parallel plate capacitor need to be in order to make the capacitance equal to 1F, if d = 8.85 mm?

C = E0A/d Cd= E0A A = Cd/E0 A = (1)(8.85 x 10^-3 m) / (8.85 x 10^-12 F/m)

Capacitance of Parallel Plate Capacitor with Dielectric

C = E0kA/d

What happens to C, Q, V, and E when the battery is connected and a dielectric is inserted?

C increases by a factor of K Q increases by a factor of K V stays the same E stays the same

What happens to C, Q, V, and E when the battery is disconnected and a dielectric is inserted?

C increases by a factor of K (always increases) Q stays the same V decreases by a factor of K E decreases by a factor of K

Two copper wires are tested for resistance. The first wire is longer and thinner than the second wire. Which wire has the higher resistivity? A. The first wire, because resistance increases with increased length and decreased area. B. The second wire, because resistance increases with decreased length and increased area. C. Neither wire, because resistivity depends only on the material, and both wires are made of copper. D. It cannot be determined without knowing the voltage and current, because Ohm's law defines resistance as a ratio of voltage over current.

C; Resistivity is a measure of a material's intrinsic resistance, and is dependent only on the material itself. Since both wires are made of copper, the resistivity of each wire will be the same. Note that answer choice A would be correct if the question were asking for resistance. Note that answer choice D correctly states Ohm's law, but is incorrect because resistance is determined by the wire and is fixed for Ohmic materials, regardless of the voltage and current.

A test charge q has a velocity of 40 m/s at point A (electric potential of 300 V). The charge slows to a stop at point B (electric potential of -100 V). If the mass of the charge is 10-6 kg, what is the charge on q? A. 2 × 10-6 C B. 4 × 10-6 C C. -2 × 10-6 C D. -4 × 10-6 C

C; Since the charge begins with a velocity and slows to a stop, the velocity and final kinetic energy are zero. The change in kinetic energy is all the initial kinetic energy so KE = (1/2) mv2 = (1/2)(10-6)(40)2 = 8 × 10-4. This loss in kinetic energy is the same as the gain in potential energy. The gain in potential energy is ΔPE = qV and solving for q yields q = ΔPE / V = 8 × 10-4 / (-100 - 300) = -2 × 10-6 C. Note that the charge must be negative since it is losing kinetic energy and gaining potential energy while moving to a lower potential

A portion of a circuit is shown below: If the current through the 10 ohm resistor is 1A, what is the current through a 20 ohm resistor? It is a circuit in series, R1 = 10 ohm, R2 = 40 ohm, R3 = 20 ohm, and the current through R1 = 1A A. .25A B. .5A C. 1A D. 2A

C; The resistors are in series, so current is constant

Capacitors in Parallel Equation

C=C₁+C₂+C₃+...

Capacitance on a parallel plate capacitor

C=EoA/d C = capacitance (F, or C/V) E0 = Permitivity of free space (8.85 x 10^-12 F/m) A = area (m^2) d = distance between plates (m)

Capacitance equation

C=Q/V C = capacitance (F, or C/V) Q = charge (C) V = voltage (V, or J/C)

A proton (whose mass is m) is placed on top of the positively charged plate of a parallel plate capacitor, as shown below. The charge on the capacitor is Q, and the capacitance is C. If the electric field in the region between the plates has a magnitude E, which of the following expression gives the time required for the proton to move up to the other plate? A. d(sqrt eQ/mC) B. d(sqrt m/eQC) C. d(sqrt 2eQ/mC) D. d(sqrt 2mC/eQ)

D

How much energy is dissipated in 10 seconds by the 24 ohm resistor in the following circuit? V = 64 V R1 and R2 are in parallel with each other, R1 = 8 ohm, R2 = 24 ohm R3 and R4 are in parallel, R3 = 4 ohm and R4 = 4 ohm A. 480 J B. 640 J C. 720 J D. 960 J

D R1 * R2 / (R1 + R2) = (8)(24) / 32 = 96/16 = 6 ohms R3 * R4 / (R3 + R4) = 16/8 = 2 ohms Series: 2 ohm + 6 ohm = 8 ohm V = IR I = V/R I = 64V/8 ohm = 8A 3 times as much current goes through 2 ohm resistor than 6 ohm resistor V = (2A)(24 ohm) V = 48 V V = (6A)(8 ohm) V = 48 V P = I^2R P = (4)(24) P = 96W P = W/t W = (P)(t) W = (96W)(10s) W = 960 J

Jeremiah shuffles his stocking feet on the carpet, accumulating a charge of -10 nC, and then reaches for the doorknob. When his finger approaches the metal, a spark flies between his finger and the doorknob. If the spark lasts 1 microsecond, what is the instantaneous current of the spark? A. 5 nA B. 10 nA C. 5 mA D. 10 mA

D; Current is defined as charge per time, or Q / t. In this case, 10 × 10-9 / 1 × 10-6 = 10 × 10-3 A = 10 mA. Note that even though the charge is negative, current is always defined as the positive flow of charge.

Wire X is made of copper and has a length L and a cross-sectional area A. Wire Y is also made of copper and has twice the length and half the radius of Wire X. What is the ratio of the resistance of Wire X to the resistance of Wire Y? A. 1 : 1 B. 1 : 2 C. 1 : 4 D. 1 : 8

D; Since both wires are made of copper, they will have the same resistivity, so the resistance is proportional to L / A or L / r2 where r is the radius. Wire Y has twice the length and half the radius so the proportion becomes 2L / (0.5r)2or 8 L / r2. Thus the ratio of the resistance of Wire X to Wire Y is 1 to 8.

All of the following will increase the amount of charge able to be stored on a capacitor EXCEPT: A. Increase the area of the plates on the capacitor B. Decrease the distance between the plates on the capacitor C. Increase the voltage of the battery used to charge the capacitor D. Increase the resistance in the circuit used to charge the capacitor

D; The charge stored on a capacitor is given by Q = CV where C is the capacitance and V is the voltage difference on the capacitor. So anything that increases the capacitance or the voltage on the capacitor, will increase the charge stored. The capacitance is directly proportional to the area of the plates, eliminating choice A. The capacitance is indirectly proportional to the distance between the plates, eliminating choice B. Using a battery with a larger voltage will allow the capacitor to charge to a larger voltage, which would increase the charge stored, eliminating choice C. Increasing the resistance in the circuit will not modify the charge stored on the capacitor, making choice D false and the correct answer.

A student in physics lab creates an electric field using a parallel plate capacitor. The top plate has a positive charge and the bottom plate has a negative charge, creating a uniform electric field between the plates. The student is testing multiple particles of varying mass m and charge q. The particles are fired into the capacitor halfway between the plates with a velocity v parallel to the plates. Then the particles are collected and their velocities recorded as they exit the capacitor. Which of the following best explains why some of the particles do not exit the capacitor to be recorded? A. The electric force on the particles with larger charge is so great that the particles stop moving and remain between the capacitor plates. B. The electric force on negative particles moves them directly to the top capacitor plate so that they cannot ever exit the capacitor. C. The electric force on particles with larger mass is not strong enough to move the particles through the capacitor. D. The electric force on the particles causes them to move parabolically, and since particles with larger charge will curve faster, they may not exit the capacitor before colliding with one of the plates.

D; The parallel plate capacitor creates a uniform electric field E with field lines that point from the positive plate to the negative plate. The electric field will exert a force F = qE on a charged particle q passing into the electric field. This force creates an acceleration in the direction of the force (up for negative charges, down for positive charges). Since the force and acceleration are constant in magnitude and are perpendicular to the plates, and because the particle has an initial velocity parallel to the plates, the particle will travel parabolically (not unlike an object experiencing projectile motion accelerating due to gravity). If the charge is larger, the force and acceleration will be larger, the curve will be tighter, and the particle will collide with the capacitor wall before exiting the capacitor. Note that choice A is false since the electric force will cause the particles to accelerate (change direction), not stop moving. Choice B is false because a particle's initial velocity parallel to the plates causes it to move in a curve (additionally, choice B implies ALL particles do not exit, while the question states only some do not exit). Choice C is false because the initial velocities of the particles is what causes them to pass between the plates, not the electric force.

Voltage that creates a current is referred to as

Electromotive force

What are resistors in series?

Follow each other along a single connection

Current equation?

I = Q/t I = current (A) Q = charge (C) t = time (s)

A typical ion channel in a cellular membrane might allow the passage of 10^7 sodium ions to flow through in one second. What is the magnitude and direction of this ionic current?

I = Q/t I = (10^7 Na+)(1.6 x 10^-19 C) / 1s I = 1.6 x 10^-12 A

Within a metal wire, 5 x 10^17 conduction electrons drift pas a certain point in 4 seconds. What is the magnitude of the current?

I = Q/t I = (5 x 10^-17 e-)(1.6 x 10^-19 C) / 4s I = 2 x10 ^-2 or .02A

Are batteries a source of current?

No

A charged capacitor has a charge of Q, and the voltage between the plates is V. What will happen to C if Q is doubled?

Nothing; C is constant for a given capacitor

Describe a parallel plate capacitor

One plate carries a positive charge Other plate carries a negative charge Net charge is 0 Electric field points from positive plate to negative plate

Power supplied to the circuit by a voltage source

P = IV

Power dissipated by a resistor equation

P = I^2R

A toaster oven is rated at 720 W. If it draws 6A of current, what is its resistance?

P = I^2R R = P/I^2 R = 720 W / 36 R = 20 ohm

Electric potential energy equations

PE = 1/2QV PE = 1/2CV^2 PE = Q^2/2C Q = charge (C) C = capacitance (F, or C/V) V = potential difference (V, or J/C)

What are resistors in parallel?

Provide alternate routes from one point in a circuit to another.

Charge on a capacitor equation

Q = CV Q = charge (C) V = potential difference (V) C = capacitance (F, or C/V)

The charge on a parallel plate capacitor is 4 x 10^-6C. If the distance between the plates is 2mm and the capicitance is 1 uF, what is the strength of the electric field between the plates?

Q = CV V = Q/C V = 4 x 10^-6/1 x 10^-6 V = 4V V = Ed E = V/d E = 4V / 2 x 10^-3 m E = 2 x 10^3 V/m

Electric potential energy is proportional to what?

Q and V

The wire sued for lighting systems is usually No. 12 wire, in the American Wire Gauge (AWG) system. The diameter of No. 12 wire is just over 2 mm (which means the cross sectional area of 3.3 x 10^-6 m^2). What would be the resistance of half a mile (800m) of No.12 copper wire, given that the resistivity of copper is 1.7 x 10^-8 Ohm * m?

R = pL/A R = (1.7 x 10^-8 ohm/meter)(800 ohm) / (3.3 x 10-6 m^2) R = 4 ohm

Resistivity equation

R = ρL/A R, resistance / Ω ρ, resistivity / Ωm L, length / m A, cross section area / m2

Resistance equation

R=V/I R = resistance (ohms, or V/A) V = voltage (volts) I = current

Equivalent resistance in series

Rs=R1+R2+R3+...+Rn

Show that the equivalence of two identical resistors in series is twice the resistance of either resistor, but the equivalent resistance of two identical resistors in parallel is half the resistance

Series: Req R + R = 2R Parallel: Req = (R * R) / (R+R) = R^2/2R = R/2

What is a dielectric?

Slab of insulating material placed between plates of a capacitor

What are conduction electrons?

The loosely bound electrons of the outermost orbit which leave their individual atom and become free to move inside the solid.

What is the current in the 100 ohm resistor shown below? V = 10 V R1 and R2 are in parallel, R2 and R3 are in parallel

The parallel combination is attached directly to the positive terminals of the battery. V = IR I = 10V/100 ohm = .1A

What is current?

The rate of flow of charge

What is Kirchhoff's loop rule?

The sum of the voltage sources (batteries) must be equal to the sum of the voltage (potential) drops

What is Kirchoff's junction rule?

The total current flowing into a junction must equal the total current flowing out of it.

What causes current? What would make an electron drift to the right?

There is a potential different (a voltage) between the ends of the wires. Negative charges move toward regions of higher electric potential.

What direction does the electric field point in a capacitor?

Toward the negative plate

The plates of a parallel plate capacitor are separated by a distance of 2mm. The device's capacitance is 1 uF. How much charge needs to be transferred from one plate to another in order to create a uniform electric field whose strength is 10^4 V/m

V = ED V = (10^4)(2 x 10^-3 m) V = 20V Q = CV Q = (1 x 10^-6 F)(2 x 10^1) Q = 2 x 10^-5 C

Ed's formula

V = Ed V = electric potential E = electric field d = separation of plates

The battery in the circuit below has an emf of 100V and an internal resistance of 5 ohms. What is the terminal voltage in the circuit? (Note: It's not uncommon to see a dashed box drawn around the battery and its internal resistance: this emphasized that r is actually inside the battery R1/R2/R3 = in series R1 = 5 ohm R2 = 20 ohm R3 = 25 ohm

V = IR I = V/R I = 100V / 50 ohm I = 2A V = E - IR V = 100 V - (2)(5) V = 90V

When the potential difference between the ends of a wire is 12V, the current is measured to be .06A. What is the resistance of the wire?

V = IR R = V/I R = 12 V / 6 x 10^-2 A R = 200 ohm

Terminal voltage equation

V = emf - Ir I = current r = resistance

What is the current in the circuit below? V1 = 100 V in the right direction V2 = 10 V in the left direction R1/R2/R3 are in series R1 = 30 ohm, R2 = 20 ohm, R3 = 10 ohm

V(total) = V1 - V2 V(total) = 90 V Req = R1 + R2 + R3 Req = 60 ohm V = IR I = V/R I = 90V/60 ohm = 1.5 A

What is Ohm's Law?

V=IR V = voltage (V) I = current (A) R = resistance (ohm, V/A)

A capacitor is connected in a circuit to a battery and allowed to fully charge. While the capacitor is still connected to the battery, a dielectric is inserted between the plates of the capacitor. If the dielectric constant is K, all of the following are true EXCEPT: A. The charge stored on the capacitor increases by a factor of K. B. The potential energy stored on the capacitor increases by a factor of K. C. The voltage across the capacitor increases by a factor of K. D. The electric field created by the capacitor remains the same.

When the dielectric is inserted, the capacitance of the capacitor increases by a factor of K. Since the battery is still connected, the capacitor will continue charging until the voltage on the capacitor is the same as the voltage on the battery. Therefore V will remain the same, making choice C false and the correct answer. Using Q = CV with V the same and C increasing by a factor of K, then Q must also increase by a factor of K, eliminating choice A. Using PE = (1/2) QV then PE must also increase by a factor of K, eliminating choice B. Using V = Ed with V the same, then E must also be the same, eliminating choice D.

What is a resistor?

a component of an electrical circuit that has resistance and is used to control the flow of electric current

What is a capacitor?

a component that stores charge

What is the dielectric constant?

a quantity measuring the ability of a substance to store electrical energy in an electric field

SI unit of current

ampere (1 A = 1 coulomb/second)

Ammeter measures

current

Series resistors have the same _______

current

What do dielectric materials do to capacitance?

increase it

Ideal resistance of a voltmeter

infinite

Ideal resistance of an ohmmeters

infinity

All real batteries have _________ resistance

internal

The terminal that is at the lower potential is denoted by the shorter line and is called the ________ terminal

negative

placement of a voltmeter

parallel

Conduction electrons will flow towards the _________ terminal

positive

The terminal that is at a higher potential is denoted by the longer line and is called the ________ terminal

positive

A capacitor is a storage device for electric _________ energy

potential

Voltmeter measures

potential difference

Ohmmeter measures

resistance

The simplest circuit consists of a voltage source, a connecting wire between terminals and a _______

resistor

How is an ammeter placed in a circuit?

series

Ohmmeters are placed in

series

Capacitors in series all have the same

to charge

For a capacitor to be useful, the plates cannot what?

touch

Capacitors in parallel have the same

voltage

Parallel resistors have the same ______

voltage

What is resistance?

Anything that slows the flow down

Units of capacitance

Farads = 1 C/V